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Math Help - Derive Logistic Growth Equation

  1. #1
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    Derive Logistic Growth Equation

    Hey guys. I'm really stuck on this homework. The instructions were "Dervie the logistic growth equation".

    So, the teacher gave us the following stuff:

    dy/dt = ky (M-y)
    k, t > 0

    y(0) = M / 1+c

    y = M / 1+ce ^ -kMt

    So, i'm guessing I need to dervie y = M / 1+ce ^ -kMt right?

    This is how i went about it.

    y = M / 1+ce ^ -kMt
    dy/dt = (1+ce ^ -kMt)(M derivative) - (M)(-kMt ce ^ -kMt-1)(ce derivative) / (1+ce ^ -kMt)^2

    Not sure if i'm doing the correct thing though. Please help me out!
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  2. #2
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    Lexington, MA (USA)
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    Hello, anfieldgirl!

    This is a Differential Equation problem.
    . . We must solve it accordingly.


    Derive the logistic growth equation.

    . . \frac{dy}{dt} \:=\: ky (M-y),\;\;k, t > 0,\;\;y(0) \:=\:\frac{M}{1+c}

    Answer: . y \:= \:\frac{M}{1+ce^{-kMt}}

    We have: . \frac{dy}{dt}  \:=\:ky(M-y)

    Separate variables: . \frac{dy}{y(M-y)} \:=\:k\,dt


    On the left, use Partial Fraction Decomposition:

    . . \frac{1}{M}\left(\frac{1}{y} + \frac{1}{M-y}\right)dy \:=\:k\,dt


    Integrate: . \frac{1}{M}\left[\ln y - \ln(M-y)\right] \;=\;kt + C_1 \quad\Rightarrow\quad \ln\left(\frac{y}{M-y}\right) \:=\:kMt + C_2

    . . Then: . \frac{y}{M-y} \:=\:e^{kMt + C_2} \:=\:e^{kMt}\cdot e^{C_2} \;=\;Ce^{kMt}


    We have: . y \:=\:Ce^{kMt}(M - y) \quad\Rightarrow\quad y \;=\;CMe^{kMt} - Cye^{kMt}

    . . Cye^{kMt}+ y \;=\;CMe^{kMt} \quad\Rightarrow\quad y\left(Ce^{kMt} + 1\right) \;=\;CMe^{kMt}<br />

    Then: . y \;=\;\frac{CMe^{kMt}}{C^{kMt}+1}


    Divide top and bottom by e^{kMt}\!:\;\;y \;=\;\frac{CM}{e^{-kMt} + 1}

    Then use the initial condition to evaluate C.

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  3. #3
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    Thanks so much for replying. I have some questions, i need to clarify though.

    First, why do we seperate the variables?

    Second, is there a different way other than Partial Fraction Decomposition? Because my teacher didn't teach us that method yet, so i'm confused...

    Thank you so much again.
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